Question

The line $3x+2y=13$ divides the area enclosed by the curve $9x^2+4y^2-18x-16y-11=0$ in two parts Find the ratio of the larger area to the smaller area

• A. $\frac{3\pi +2}{\pi -2}$
• B. $\frac{3\pi -2}{\pi +2}$
• C. $\frac{\pi +2}{\pi -2}$
• D. $\frac{\pi -2}{\pi +2}$

Answer 1

The correct option is A $\frac{3\pi +2}{\pi -2}$

$9x^2+4y^2-18x-16y-11=0\Rightarrow 9(x^2-2x)+4(y^2-4y)=11\Rightarrow 9({(x-1)}^2-1)+4({(y-2)}^2-4)=11\Rightarrow 9{(x-1)}^2+4{(y-2)}^2=36$
$\Rightarrow \frac{{(x-1)}^2}{4}+\frac{{(y-2)}^2}{9}=1$
Let $x-1=X$ and $y-2=Y$
$\Rightarrow \frac{X^2}{4}+\frac{Y^2}{9}=1$
Hence $3x+2y=13$
$\Rightarrow 3(X+1)+2(Y-2)=13\Rightarrow 3X+2Y=6\Rightarrow \frac{X}{2}+\frac{Y}{3}=1$
Therefore area of triangle $OPQ=\frac{1}{2}\times 2\times 3=3$
Also area of ellipse $=\pi$(semimajor axes)(semi minor axis) $=\pi .3.2=6\pi$
$A_1=\frac{6\pi }{4}-$ area of $△OPQ=\frac{3\pi }{2}-3$
$A_2=3\left(\frac{6\pi }{4}\right)+$ area of $△OPQ=\frac{9\pi }{2}+3$
Hence $\frac{A_2}{A_1}=\frac{\frac{9\pi }{2}+3}{\frac{3\pi }{2}-3}=\frac{3\pi +2}{\pi -2}$